Các câu hỏi tương tự
1, Cho
A = \(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{2005.2006}\)
B = \(\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.......+\frac{1}{2006.1004}\)
Hãy Tính \(\frac{A}{B}\)
Cho:
\(A=\frac{1}{1.2}+\frac{1}{3.4}+.....+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+.....+\frac{1}{2006.1004}\)
Hãy tính \(\frac{A}{B}\)
Cho
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
Tìm \(\frac{A}{B}\)
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Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+............+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+.....+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
A=1/1.2+1/3.4+...+1/2005.2006
B=1/1004.2006+1/1005.12005+...+1/2006.1004
Chung minh A/B la so nguyen
\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2003.2004}+\frac{1}{2005.2006}\)
\(B=\frac{1}{1004.2006}+\frac{1}{1005.2005}+\frac{1}{1006.2004}+...+\frac{1}{2006.1004}\)
Tính \(\frac{A}{B}\)
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Cho A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+\(\frac{1}{5.6}\)+...+\(\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}\)+\(\frac{1}{1005.2005}\)+...+\(\frac{1}{2006.1004}\)
Chứng minh \(\frac{A}{B}\) là số nguyên
Bài 1 : Tính hợp lý :a) A 182 . left(frac{1+frac{1}{3}+frac{1}{9}+frac{1}{27}}{2+frac{2}{3}+frac{2}{9}+frac{2}{27}}:frac{4-frac{4}{7}+frac{4}{49}-frac{4}{343}}{1-frac{1}{7}+frac{1}{49}-frac{1}{343}}right) : frac{919191}{808080}b) C -2+frac{1}{-2+frac{1}{-2+frac{1}{-2+frac{1}{3}}}}c) D left(frac{1}{1.2}+frac{1}{3.4}+...+frac{1}{2005.2006}right): left(frac{1}{1004.2006}+frac{1}{1005.2005}+...+frac{1}{2006.1004}right)
Đọc tiếp
Bài 1 : Tính hợp lý :
a) A = 182 . \(\left(\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right)\) : \(\frac{919191}{808080}\)
b) C = \(-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{-2+\frac{1}{3}}}}\)
c) D = \(\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2005.2006}\right)\): \(\left(\frac{1}{1004.2006}+\frac{1}{1005.2005}+...+\frac{1}{2006.1004}\right)\)
1;Tính:
A=22+42+62+82+...+1002.
B=13+23+33+43+...+1003.
2;Cho A=\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
B=\(\frac{1}{1004.2006}+\frac{1}{1005.2006}+...+\frac{1}{2006.2006}\)
Tính A : B